Is a mixed second rank tensor reducible?

[jason12345]

As a complete novice, I'm reading a text which says that a mixed second rank tensor T^{u}_{v} is reducible but don't see how. Anyone care to show me?

 

[robphy]

What is the definition of reducible? (I'm not sure what you mean.)

 

[jason12345]

What is the definition of reducible? (I'm not sure what you mean.)

It means it can be broken down into parts that transform among themselves.

 

[robphy]

Can you provide the reference where the statement in your first post appears?

 

[jason12345]

Can you provide the reference where the statement in your first post appears?

It comes from Anderson's Principles of Relativty Physics on page 19:

" Thus from the components of a tensor T^{uv} we can construct its symmetric part T^{(uv)} and its antisymmetric part T^{[uv]} according to

T^{(uv)} = 1/2(T^{uv}+ T^{vu})

T^{[uv]} = 1/2(T^{uv}- T^{vu})

Similarly we can construct the transformed symmetric part T^{'(uv)} and the antisymmetric part T^{'[uv]} from the transformed T^{'uv}. Then, one can show that T^{'(uv)} is a function of T^{(uv)} and the mapping function only, and similarly for T^{[uv]}. When ever a geometrical object can be broken up into parts that transform among themselves, we say that we have a reducible object. If no such decomposition as possible. we have an irreducible object.

 

[robphy]

OK, I see now.
I take it you are referring to Problem 1.7...
where they are telling you to "decompose" T_\nu{}^\mu
into irreducible parts consisting of
its "trace" T_\mu{}^\mu and
a "traceless " (or "trace-free") tensor (which they give as T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho on the assumption one is working in a 4-dimensional space).

So, basically they have told you to write T_\nu{}^\mu
as a sum of two tensors, one of which they gave you:
T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu

 

[jason12345]

OK, I see now.
I take it you are referring to Problem 1.7...
where they are telling you to "decompose" T_\nu{}^\mu
into irreducible parts consisting of
its "trace" T_\mu{}^\mu and
a "traceless " (or "trace-free") tensor (which they give as T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho on the assumption one is working in a 4-dimensional space).

So, basically they have told you to write T_\nu{}^\mu
as a sum of two tensors, one of which they gave you:
T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu

Yes. The question says: "show that it is reducible *and* its irreducuble parts are...". So I would have thought you should be able to first show its reducible from the transformation properties of a mixed second rank tensor, rather than being told what the irreducible components are and from that showing its reducible :smile